hi there

i just started an algebra class and my teacher gave us this problem to do for homework and I need some help on working though it:

Find all solutions to the system \begin{align*} a + b &= 14, \\ a^3 + b^3 &= 812. \end{align*}

tysm in advance!! :D

Guest Feb 15, 2022

#1**0 **

from equation 1 we get b = 14-a

substituting that into the second equation we get a^3+(14-a)^3=812

now you can try to expand it out, if you need help just ask

Guest Feb 15, 2022

#2**+2 **

So a^3 + b^3 = (a + b)(a^2 - ab + b^2).

Substituting in, we have 812 = 14(a^2 - ab + b^2).

Simplifying we have 58 = a^2 - ab + b^2.

a^2 - ab + b^2 = (a + b)(a + b) - 3ab.

Substituting in, we have 58 = (14)(14) - 3ab.

Now we have ab = 46.

Then we also have a^2 - ab + b^2 = (a - b)(a - b) + ab.

Thus, 58 = (a - b)^2 + 46.

(a - b)^2 = 12.

\(a - b = \pm{2\sqrt{3}}\)

Now we can sove for a. If we add the two equations together, we get 2a = 14 + 2sqrt(3)

Thus, \(a = 7 + \sqrt{3}\), and \(a = 7 - \sqrt{3}\).

Substituting in, we have \(b = 7 - \sqrt{3}\), and \(b = 7 - 3\sqrt{3}\).

**Thus, our possible for solutions \((a, b)\)are:**

\((7 + \sqrt{3}, 7 - \sqrt{3})\)

\((7 - \sqrt{3}, 7 - 3\sqrt{3})\)

proyaop Feb 15, 2022